---
title: "L-L fault on 40 MVA generator"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:ll-fault-40mva-generator"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-40mva-generator
site: BCT Engineering
---

# L-L fault on 40 MVA generator

Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems

- Asked in: 2078 Kartik, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/ll-fault-40mva-generator
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A 40 MVA, 11 kV generator has Z1 = Z2 = j0.3 pu, Z0 = j0.4 pu. A line to line fault occurs on the generator terminals. Find the fault current.

## Short answer

Positive and negative sequence networks parallel, Ia1 = -Ia2 = 1/(j0.6) = -j1.667 pu. Fault current magnitude in phases b and c = sqrt(3)*|Ia1| = 2.887 pu = 6.06 kA.

## Answer

**Data:** 40 MVA, 11 kV, $Z_1 = Z_2 = j0.3$ pu, $Z_0 = j0.4$ pu. Generator unloaded, $E_a = 1\angle0^\circ$ pu. Fault between phases b and c.

#### Fault conditions
$I_a = 0,\ I_b = -I_c,\ V_b = V_c$, which give $I_{a0} = 0$ and $I_{a2} = -I_{a1}$: the positive and negative sequence networks are connected in parallel (opposing), and $Z_0$ does not appear.

#### Sequence currents
$$
I_{a1} = -I_{a2} = \frac{E_a}{Z_1 + Z_2} = \frac{1}{j0.6} = -j1.667\ \text{pu}
$$

#### Fault current
$$
\begin{aligned}
I_b &= (a^2 - a)I_{a1} = -j\sqrt3\, I_{a1} = -j\sqrt3(-j1.667) = -2.887\ \text{pu} \\
I_c &= -I_b = 2.887\ \text{pu}
\end{aligned}
$$
Base current:
$$
I_{base} = \frac{40\times10^6}{\sqrt3 \times 11\times10^3} = 2099.5\ \text{A}
$$
$$
|I_f| = 2.887 \times 2099.5 = 6061\ \text{A}
$$

**Answer: L-L fault current = 2.887 pu ≈ 6.06 kA** (equal and opposite in phases b and c; $I_a = 0$).

For comparison, a 3-phase fault would give $1/0.3 = 3.333$ pu, so $I_{LL} = \frac{\sqrt3}{2} I_{3\phi}$ when $Z_1 = Z_2$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is Z0 = j0.4 not used?
   A. Because it is too large
   B. Because Ia0 = 0 for an L-L fault
   C. Because the generator is ungrounded
   D. Because Z0 equals Z1
   Answer: B. An L-L fault has no zero sequence current, so the zero sequence network plays no part. (Key point: Zero sequence absent in L-L)

2. What is Ia1?
   A. 1/j0.3 = -j3.333 pu
   B. 1/j1.0 = -j1 pu
   C. 1/j0.6 = -j1.667 pu
   D. 1/j0.7 = -j1.43 pu
   Answer: C. Ia1 = Ea/(Z1 + Z2) = 1/j0.6 = -j1.667 pu. (Key point: Positive sequence current)

3. What is the L-L fault current in pu and in kA?
   A. 1.667 pu = 3.5 kA
   B. 2.887 pu = 6.06 kA
   C. 3.333 pu = 7.0 kA
   D. 5.0 pu = 10.5 kA
   Answer: B. |Ib| = sqrt3 x 1.667 = 2.887 pu; base current is 2099.5 A so 6061 A. (Key point: Fault current magnitude)

4. When Z1 = Z2, how is the L-L current related to the 3-phase fault current?
   A. Equal
   B. Twice
   C. sqrt3/2 times
   D. 1/sqrt3 times
   Answer: C. I_LL = sqrt3/(2Z1) = (sqrt3/2) x I_3phase; here 2.887 vs 3.333 pu. (Key point: Relation to 3-phase fault)

## Related in Power System Analysis II

- [Equivalent Z0, Z1, Z2 at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asequence-thevenin-z0-z1-z2-bus3) — Power System Analysis II, chapter 5
- [L-L fault currents and voltages, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-30mva-currents-voltages) — Power System Analysis II, chapter 5
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